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If sin cost0 1,then prove that sin. cos8ă1a3then prove that sin39, cos' Ď1(a + b)3b a +b,b3( |
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Answer» Sin⁴x/a + cos⁴x/b = 1/(a + b) we know, sin²x + cos²x = 1 ⇒cos²x = 1 - sin²xcos⁴x = (1 - sin²x)² = 1 + sin⁴x - 2sin²x , use it above sin⁴x/a + (1 + sin⁴x - 2sin²x)/b = 1/(a + b) sin⁴x/a + 1/b + sin⁴x/b - 2sin²x/b = 1/(a + b) sin⁴x(1/a + 1/b) -2sin²x/b = 1/(a + b) - 1/b sin⁴x(a + b)/ab - 2asin²x/ab = (b - a - b )/(a + b)b (a + b)²(sin²x)² - 2a(a + b) sin²x = -a²{(a + b)sin²x}² -2.a.(a + b)sin²x + a² = 0 [ this is like (a - b)² = a² - 2ab + b² ]{(a + b)sin²x - a}² = 0sin²x = a/(a + b) ⇒1 - sin²x = cos²x = 1 - a/(a + b) = b/(a + b) hence, sin²x = a/(a + b) and cos²x = b/(a + b) so, sin⁸x = a⁴/(a + b)⁴ and cos⁸x = b⁴/(a + b)⁴ now, sin⁸x/a³ = a/(a + b)⁴ cos⁸x/b³ = b/(a + b)⁴ So, sin⁸x/a³ + cos⁸x/b³ = a/(a + b)⁴ + b/(a + b)⁴ = (a + b)/(a + b)⁴ = 1/(a + b)³ Like my answer if you find it useful! |
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